Significance of D'alembert's principle

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  • #1
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Main Question or Discussion Point

Does the mathematical exp (Fa-dp/dt).dr = 0

Fa being applied force, p being momentum, dr being virtual displacement


simply mean that the mechanical system will be in equilibrium if the constraint forces do no work?

Can someone elaborate on what a dynamic equilibrium means and how it differs from static mathematically?
 

Answers and Replies

  • #2
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Well, you've got A x B = 0, so this equation means that either F_a = dp/dt OR[inclusive] dr = 0. I don't get it. F_a - dp/dt is always zero, so I don't know what significance the dr has.

You confused me too, with the abbreviation "exp" which often means exponentiate.

But honestly, that's just what I made of it. I'm sure it's an important equation, so hopefully someone will give you a better answer.
 
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  • #3
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By exp I meant expression not exponent sorry. Mazerakham, this principle apparently has a lotta significance which i'm trying to figure out. The Euler-Lagrangian equations can be derived from this.
 
  • #4
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It seems pretty well explained here:
http://en.wikipedia.org/wiki/D%27Alembert%27s_principle" [Broken]
 
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  • #5
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It seems pretty well explained here:
http://en.wikipedia.org/wiki/D%27Alembert%27s_principle" [Broken]
Well I had read that before, but I'm still unclear on the points that I raised. I don't quite understand what this "dynamic" equilibrium means.

I think that in rigid bodies it might mean that for equilibrium there must be no disfigurement of the body

And also it seems to say that one does not need to consider constraint forces to determine this dynamic equilibrium whatever it is.

But to me that seems kind of a circular logic.
I mean after all the reason I want to know about the forces acting on the body is to predict the motion of the system, but here the condition for constraint forces to be ignored is, that dp/dt or more specifically d2r/dt2 be known.

I.e If I 'know' the acceleration and the applied forces , then the constraint forces can be ignored.

But again why would I want to know the constraint forces unless i'm trying to determine the acceleration ( ie motion of the system )
 
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  • #6
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I'm having the same problem, and to the first part of your question:
Does the mathematical exp (Fa-dp/dt).dr = 0

Fa being applied force, p being momentum, dr being virtual displacement


simply mean that the mechanical system will be in equilibrium if the constraint forces do no work?
I think that the constraint forces never do work, even if there is no equilibrium. So:
in equilibrium its true that: sum F=0
its always true that:sum F=sum Fa + sum Fc
its always true that: sum Fc*dr=0

thus in equilibrium it must be true that: sum Fa dr=0

So it means that the mechanical system will be in equilibrium if the constraint forces do no work on virtual displacements, which is a stronger statement than sum F=0 coming from Newton.
 
  • #7
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I'm having the same problem, and to the first part of your question:


I think that the constraint forces never do work, even if there is no equilibrium. So:
in equilibrium its true that: sum F=0
its always true that:sum F=sum Fa + sum Fc
its always true that: sum Fc*dr=0

thus in equilibrium it must be true that: sum Fa dr=0

So it means that the mechanical system will be in equilibrium if the constraint forces do no work on virtual displacements, which is a stronger statement than sum F=0 coming from Newton.
Hmm so it seems to say that,

For a system to be in equilibrium,

The components of the applied forces along the direction of any arbitrary virtual displacement must sum to zero.

Since virtual displacements are be definition 'consistent' with the constraint forces ( which I can only understand as being perpendicular am I right? )

this would imply that for any direction perpendicular to the constraint forces, the components of the applied forces would sum to zero.

This holds of course only when the constraint forces do no work
 
  • #8
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I'm not sure what you mean by "along the direction" and "for any direction". If you mean the dot product of the forces and the virtual displacements, then yes. But constraint forces never do any work. A constraint force is like an inpenetrable wall, if its penetrated then its not an inpenetrable wall anymore.
 
  • #9
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I'm not sure what you mean by "along the direction" and "for any direction". If you mean the dot product of the forces and the virtual displacements, then yes. But constraint forces never do any work. A constraint force is like an inpenetrable wall, if its penetrated then its not an inpenetrable wall anymore.
Yes the dot product of the virtual displacement and the applied forces is what I meant, but when I said it should sum to zero, I was thinking of a static equilibrium.

Just for clarification, am I to understand that the D'Alembert's principle

(Fa-dp/dt).dr

will hold for all cases where the constraint forces do no work ie they are impenetrable as you say.

If this is so can not dynamic equilibrium be thought of as a structure which undergoes *reversible* motion as long as the constraint forces stay impenetrable
 
  • #10
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D' Alembert's principle is a convenient way of preserving the Euler-Lagrange equations when the configuration space assumes a manifold structure.

If the configuration space is a physical surface, D' Alembert's principle is equivalent to the statement that "the constraint force is perpendicular to the surface." For more complex systems. D' Alembert's principle doesn't have a realistic (intuitive) meaning.

As far as "virtual displacement," it is an archaic term. We say that the virtual displacement is an element of the tangent space of the configuration manifold. D' Alembert's principle can then be summed up by saying that

"The work done by the constraint force is zero." (i.e. frictionless)
 
  • #11
Pinu 7, can i also put it as "Constraints forces dessipate no energy"?
 

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